Synthetic infinite product

Let $P_{L}$ denote the finite form of the infinite product above Then we have, $P_{L} = \prod_{i=1}^{L} (1 + \frac{1}{a_{i}}) = \prod_{i=1}^{L} \frac{( a_{i} + 1)}{(a_{i})}$

$P_{L} ={\frac{a_{1}+1}{a_{1}}} \times {\frac{a_{2}+1}{a_{2}}} \times {\frac{a_{3}+1}{a_{3}}} .......... \times {\frac{a_{L}+1}{a_{L}}}$

We will shift the terms below to the left by one place ....

$P_{L} = \frac{1}{a_{1}} \frac{a_{1}+1}{a_{2}} \times \frac{a_{2}+1}{a_{3}} \times ....... \times (a_{L}+1)$

The $a_{L}+1$ term is lonely ... So lets divide and multiply by $a_{L+1}$ we get

$P_{L} = \frac{a_{L+1}}{a_{1}} \frac{a_{1}+1}{a_{2}} \times \frac{a_{2}+1}{a_{3}} \times ....... \times \frac{a_{L}+1}{a_{L+1}}$

Now we will use the given data $a_{L+1} = L(1+a_{L})$ OR $\frac{1}{L} = \frac{1+a_{L}}{a_{L+1}}$ and $a_{1} = 1$ Hence, our previous equation turns out to be

$p_{L} = \frac{a_{L+1}}{1} ( \frac{1}{1} \times \frac{1}{2} \times \frac{1}{3} ........ \times \frac{1}{L} )$

$p_{L} = \frac{a_{L+1}}{L!}$

Now taking limit as L tends to infinity we get the value of the actual infinite product

$\lim_{L \to \infty} P_{L} = \lim_{L \to \infty} \frac{a_{L+1}}{L!}= \prod_{i=1}^{\infty}(\frac{1+a_{i}}{a_{i}})$

Evaluating the limit is not that easy, as it requires to observe a pattern in the $n^{th}$ term, considering the recurrence relation we have ....

$a_{n+1} = n + n (a_{n}) = n + n( (n-1) + (n-1)a_{n-1}) = n + n(n-1) + n( n-1) a_{n-1}$

It is easy to see that ...

$a_{n+1} = n + n(n-1) + n(n-1)(n-2) + ..... + n! (1 + a_{1})$

$a_{n+1} = n + n(n-1) + n(n-1)(n-2) + ..... + n! . 2$

now picking n! common we get

$\frac{a_{n+1}}{n!} = ( 2 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...... + \frac{1}{(n-1)!} )$

Now taking limit as n tends to infinity we get ...

$\lim_{n \to \infty} \frac{a_{n+1}}{n!} = \lim_{n \to \infty} ( 2 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ...... + \frac{1}{(n-1)!} )$

Using the series expansion for euler constant e we get the limit to be ......

$\prod_{i=1}^{\infty}(\frac{1+a_{i}}{a_{i}}) = \lim_{n \to \infty} \frac{a_{n+1}}{n!} = (1 + e)$

Where e = 2.71828 and, $[.]$ denotes the greatest integer function.

Hence the required answer is $[100.(1+e)] = 371$ is the required answer.